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CIE 9709 2026 June Paper 13 Q2

A Level / CIE / P1

CIE 9709 2026 June Paper 13 Paper · Question 2

题目

Problem

(a) Express xxx\sqrt{x} in the form xnx^n, where nn is a rational number.

(1)

(b) Hence, solve the equation

x3−9xx+8=0.x^3-9x\sqrt{x}+8=0.
(3)
题目中文翻译

(a) 将 xxx\sqrt{x} 写成 xnx^n 的形式,其中 nn 为有理数。

(1)

(b) 由此,解方程

x3−9xx+8=0.x^3-9x\sqrt{x}+8=0.
(3)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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把根号改成 1/21/2 次幂,同底数相乘时把指数相加。

答题过程

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For x≥0x\ge0, xx=x1x1/2=x3/2x\sqrt x=x^1x^{1/2}=x^{3/2}, so n=3/2n=3/2.

(b)

解法一

思路

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由 (a) 可见 x3=(x3/2)2x^3=(x^{3/2})^2,可用一个新变量把原式降成二次方程。

答题过程

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Let u=x3/2u=x^{3/2}. Then the equation becomes

u2−9u+8=(u−1)(u−8)=0.u^2-9u+8=(u-1)(u-8)=0.

Thus x3/2=1x^{3/2}=1 or 88, giving x=1x=1 or x=4x=4. Both are in the original domain x≥0x\ge0.